Rotating circular pan of liquid

1. The problem statement, all variables and given/known data
A circular pan of liquid with density [itex]\rho[/itex] is centered on a horizontal turntable rotating with angular speed [itex]\omega[/itex]. Atmospheric pressure is [itex]P_{a}[/itex]. Find expressions for (a) the pressure at the bottom of the pan and (b) the height of the liquid surface as functions of the distance [itex]r[/itex] from the axis, given that the height at the center is [itex]h_{0}[/itex].

3. The attempt at a solution
I boiled up a few equations using the above equation
[itex]E_{d}[/itex] is the energy density
at the center and at height [itex]h_0[/itex]:
[tex]E_{d} = P_{a} + \rho g h_{0}[/tex]

at the center and at the bottom:
[tex]E_{d} = P_{b}(0)=P_{a}+\rho g h_{0}[/tex]
where [itex]P_{b}(r)[/itex] is the pressure at the bottom measured r from the center

r from the center and at the surface of the liquid (height h)
[tex]E_{d}=P_{a} +\frac{1}{2}\rho r^2 \omega ^2 + \rho g h[/tex]

r from the center and at the bottom
[tex]P_{b}(r) + \frac{1}{2}\rho r^2 \omega ^2 = P_{b}(0)[/tex]

Mixing all these together out pops
[tex]P_{b}(r) = P_{a} + \rho g h_{0}-\frac{1}{2}\rho r^2 \omega ^2[/tex]
and
[tex]h=h_{0} - \frac{1}{2g}r^2 \omega ^2[/tex]
but the answer has a plus sign where I have my negative signs. Pretty sure I am wrong but would have to fudge the equations to make it work.

I made a couple of assumptions:
For a given vertical segment of water, it is all travelling at the same angular speed [itex]\omega[/itex] and the atmopsheric pressure at the different liquid heights at the surface is the same.